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# Coulomb's Law

## Any two charged particles any distance from each other will experience repulsion or attraction.

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Coulomb's Law

The Coulomb Force Law states that any two charged particles (q1,q2)\begin{align*} (q_1,q_2) \end{align*} --- with charge measured in units of Coulombs --- at a distance r\begin{align*} r \end{align*} from each other will experience a force of repulsion or attraction along the line joining them equal to:

FeWhere k=kq1q2r2=8.987×109 Nm2C2The Coulomb Force [1]The Electric Constant\begin{align*}\vec{F_e} &= \frac{kq_1q_2}{r^2} && \text{The Coulomb Force [1]}\\ \text{Where } k &= 8.987 \times 10^9 \ \mathrm{N \cdot m^2 \cdot C^{-2}} && \text{The Electric Constant}\end{align*}

This looks a lot like the Law of Universal Gravitation, which deals with attraction between objects with mass. The big difference is that while any two masses experience mutual attraction, two charges can either attract or repel each other, depending on whether the signs of their charges are alike:

Like gravitational (and all other) forces, Coulomb forces add as vectors. Thus to find the force on a charge from an arrangement of charges, one needs to find the vector sum of the force from each charge in the arrangement.

Key Equations

F=kq1q2r2kq1, q2rThe electric constant, k=8.987×109 Nm2C2.Magnitude of the charges, units of Coulombs (C)Distance between charges, m\begin{align*} F = \frac{k q_1 q_2}{r^2} \begin{cases} k & \text{The electric constant,}\ k = 8.987 \times 10^9 \ \mathrm{N \cdot m^2 C^{-2}}. \\ q_1, \ q_2 & \text{Magnitude of the charges, units of Coulombs (C)}\\ r & \text{Distance between charges, m} \end{cases}\end{align*}

#### Example

Two negatively charged spheres (one with 12μC\begin{align*}-12\mu\mathrm{C}\end{align*}; the other with 3μC\begin{align*}-3\mu\mathrm{C}\end{align*}) are 3m\begin{align*}3\mathrm{m}\end{align*} apart. Where could you place an electron so that it will be suspended in space between them with a net force of zero (for this problem we will ignore the force of repulsion between the two charges because they are held in place)?

Consider the diagram above; here rse\begin{align*} r_{s\rightarrow e} \end{align*} is the distance between the electron and the small charge, while Fse\begin{align*}\vec{F}_{s\rightarrow e} \end{align*} is the force the electron feels due to it. For the electron to be balanced in between the two charges, the forces of repulsion caused by the two charges on the electron would have to be balanced. To do this, we will set the equation for the force exerted by two charges on each other equal and solve for a distance ratio. We will denote the difference between the charges through the subscripts "s" for the smaller charge, "e" for the electron, and "l" for the larger charge.

kqsqer2se=kqlqer2el\begin{align*} \frac{kq_sq_e}{r_{s\rightarrow e}^2}=\frac{kq_lq_e}{r_{e\rightarrow l}^2} \end{align*}

Now we can cancel. The charge of the electron cancels. The constant k\begin{align*}k\end{align*} also cancels. We can then replace the large and small charges with the numbers. This leaves us with the distances. We can then manipulate the equation to produce a ratio of the distances.

3μCr2se=12μCr2elr2ser2el=12μC3μCrserel=1μC4μC=12\begin{align*} \frac{-3\mu\mathrm{C}}{r_{s\rightarrow e}^2}=\frac{-12\mu\mathrm{C}}{r_{e\rightarrow l}^2} \Rightarrow \frac{r_{s\rightarrow e}^2}{r_{e\rightarrow l}^2}=\frac{-12\mu\mathrm{C}}{-3\mu\mathrm{C}} \Rightarrow \frac{r_{s\rightarrow e}}{r_{e\rightarrow l}}=\sqrt{\frac{1\mu\mathrm{C}}{4\mu\mathrm{C}}}=\frac{1}{2} \end{align*}

Given this ratio, we know that the electron is twice as far from the large charge (12μC\begin{align*}-12\mu\mathrm{C}\end{align*}) as from the small charge (3μC\begin{align*}-3\mu\mathrm{C}\end{align*}). Given that the distance between the small and large charges is 3m\begin{align*}3\mathrm{m}\end{align*}, we can determine that the electron must be located 2m\begin{align*}2\mathrm{m}\end{align*} away from the large charge and 1m\begin{align*}1\mathrm{m}\end{align*} away from the smaller charge.

### Interactive Simulation

Electric Ice Sheet (CK-12 Simulation)

### Review

1. A suspended pith ball possessing +10 μC\begin{align*}+10 \ \mu \mathrm{C}\end{align*} of charge is placed 0.02m\begin{align*}0.02 \;\mathrm{m}\end{align*} away from a metal plate possessing 6 μC\begin{align*}-6 \ \mu \mathrm{C}\end{align*} of charge.
1. Are these objects attracted or repulsed?
2. What is the force on the negatively charged object?
3. What is the force on the positively charged object?
2. Consider the hydrogen atom. Does the electron orbit the proton due to the force of gravity or the electric force? Calculate both forces and compare them. (You may need to look up the properties of the hydrogen atom to complete this problem.)
3. Find the direction and magnitude of the force on the charge at the origin (see picture). The object at the origin has a charge of 8 μC\begin{align*}8 \ \mu \mathrm{C}\end{align*}, the object at coordinates (2m, 0)\begin{align*}(-2 \;\mathrm{m}, \ 0)\end{align*} has a charge of 12 μC\begin{align*}12 \ \mu \mathrm{C}\end{align*}, and the object at coordinates (0,4m)\begin{align*}(0, -4 \;\mathrm{m})\end{align*} has a charge of 44 μC\begin{align*}44 \ \mu \mathrm{C}\end{align*}. All distance units are in meters.
4. Two pith balls of equal and like charges are repulsed from each other as shown in the figure below. They both have a mass of 2g\begin{align*}2 \;\mathrm{g}\end{align*} and are separated by 30\begin{align*}30^\circ\end{align*}. One is hanging freely from a 0.5m\begin{align*}0.5 \;\mathrm{m}\end{align*} string, while the other, also hanging from a 0.5m\begin{align*}0.5\;\mathrm{m}\end{align*} string, is stuck like putty to the wall.
1. Draw the free body diagram for the hanging pith ball
2. Find the distance between the leftmost pith ball and the wall (this will involve working a geometry problem)
3. Find the tension in the string (Hint: use y\begin{align*}y-\end{align*}direction force balance)
4. Find the amount of charge on the pith balls (Hint: use x\begin{align*}x-\end{align*}direction force balance)

1. a. attracted b. 1350N\begin{align*}1350 \;\mathrm{N}\end{align*} c. 1350N\begin{align*}1350 \;\mathrm{N}\end{align*}
2. Fg=1.0×1047N\begin{align*}F_g = 1.0 \times 10^{-47} \;\mathrm{N}\end{align*} and Fe=2.3×108N\begin{align*}F_e = 2.3 \times 10^{-8} \;\mathrm{N}\end{align*}. The electric force is 39\begin{align*}39\end{align*} orders of magnitudes bigger.
3. 0.293N\begin{align*}0.293 \;\mathrm{N}\end{align*} and at 42.5\begin{align*}42.5^\circ\end{align*}
4. b. 0.25m\begin{align*}0.25\mathrm{m}\end{align*} c. FT=0.022N\begin{align*}F_T = 0.022\;\mathrm{N}\end{align*} d. 0.37μC\begin{align*}0.37 \mu \mathrm{C}\end{align*}

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